# Natural Transformations

“I didn’t invent categories to study functors; I invented them to study natural transformations.” – Saunders Mac Lane, one of the founders of category theory

A natural transformation is, loosely speaking, a homomorphism between functors. Its definition may seem strange but the example after it should illustrate its use.

DefinitionLet be two covariant functors. A

natural transformationassigns, to each object , a morphism such that for all morphisms in , we have

In diagram, we have the following, where the right square commutes.

*In summary a natural transformation from one functor to another takes an object in the base category to a morphism between the target objects.*

### Example

Let and be functors given by

- is the group of invertible 2 × 2 matrices with entries in
*B*, - is the group of units in
*B*.

Then there is a natural transformation : for each *A*-algebra *B* we take:

Naturality means if is a homomorphism of *A*-algebras, then the following commutes:

which is easily verified.

### Exercise A

Suppose we have a morphism in the category . From *f* define a natural transformation between the functors

# Composing Natural Transformations

Definition.We can compose natural transformations as follows. For covariant functors:

,

suppose we have natural transformations and . We can compose by

If we can find natural transformations and such that

,

then T and U are called

natural isomorphisms.Functors are said to be

naturally isomorphicif there exist natural isomorphisms between them as above.

### Example 1

Take functors as above. We defined the natural transformation

which takes the determinant of the matrix. For each integer *n* we also have the following natural transformation

Composing then takes each *A*-algebra *B* to the map

Even more simply, we have .

### Example 2

Let be the category of finite-dimensional vector spaces over a fixed field *k*; the morphisms are the linear maps. Consider the functors:

,

where is the identity functor and *F* takes the vector space *V* to its double-dual (recall that is the space of all linear maps ). Each linear map gives and thus .

It follows from linear algebra that we have a natural isomorphism

One can visualize this as taking inner product for and . Fixing gives a linear while fixing gives a linear , i.e. an element of .

Hence we get a natural isomorphism which takes to .

### Optional Exercise

Define another composition of natural transformations as follows. Suppose and are covariant functors. Suppose also and are natural transformations. Define a natural transformation

Hint: in diagram form we have:

# Examples

Earlier, we defined an isomorphism

where *U*(*B*) returns the set of units of *B*. We can now prove this is natural using category theory. Note that both sides are functors in terms of *B*. We will define a natural transformation :

To prove that this is a natural isomorphism, we can define a reverse natural transformation . However, there is a simpler way. We note that for each *A*-algebra *B*, the map is bijective.

- Indeed, any invertible element gives us an
*A*-algebra homomorphism by taking to respectively.

Now apply the following to complete the proof.

Proposition 1.Suppose are functors and is a natural transformation. If, for each , the morphism in is an isomorphism, then is a natural isomorphism.

**Proof**

We need to find an inverse natural transformation . For each , let be the inverse of . To check that *T’* is natural:

- Indeed, for any in , we have .
- Since we have , which is exactly what we want. ♦

# Representable Functors

Thus, we have shown that the functor which takes *C* to its set of units can be represented by a hom functor for some *A*-algebra *B*.

Definition.Let be a category and be a (covariant) functor. We say F is

representableif there is an object such thatas a natural isomorphism.

For the case of a contravariant , we require to satisfy .

**Exercise B**

Prove that for a representable *F*, the object *A* is unique up to isomorphism.

It turns out a large class of functors are representable.

1. Consider the forgetful functor . This is representable, for we can take the the group . The natural transformation then takes a group homomorphism to the image . Clearly is a bijection of sets.

As an exercise, show that for each of the following categories, the forgetful functor which takes an object to its underlying set (and morphism to its underlying function), is representable.

What about the category of non-commutative rings?

2. Consider ,

Note that this is not a subgroup in general. If is a group homomorphism and , then so so *F* is indeed a functor. *F* is representable since

.

3. Consider which takes an *A*-algebra *B* and returns the set . *F* is representable since

.

### Optional Note

Let be the closed interval in . The functor takes a space *X* to the set of all paths in *X*. We call this the **path space** of *X*. Similarly, if is the unit circle, takes *X* to the set of all loops in *X*; this is called the **loop space** of *X*. To induce a canonical topology on these spaces, topologists prefer to work in a rather unusual category of topological spaces, those which are *compactly generated and weakly Hausdorff*, which allows them to define a very nice topology for function spaces. But that’s another story for another day.